A vessel contains an ideal monoatomic gas which expands at constant pressure, when heat Q is given to it. Then the work done in expansion is

Options

A. `Q`
B. `(3)/(5) Q`
C. `(2)/(5) Q`
D. `(2)/(3) Q`

Explanation

To find the work done in the expansion of an ideal monoatomic gas at constant pressure when heat \( Q \) is given to it, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**: We have an ideal monoatomic gas expanding at constant pressure. The heat \( Q \) is added to the system.

2. **Determine the Heat Capacity**: For a monoatomic gas, the molar heat capacities are defined as:
- \( C_v = \frac{3}{2} R \) (where \( R \) is the universal gas constant)
- \( C_p = C_v + R = \frac{3}{2} R + R = \frac{5}{2} R \)

3. **Relate Heat to Temperature Change**: The heat added at constant pressure can be expressed as:
\[ Q = n C_p \Delta T \]
Substituting \( C_p \):
\[ Q = n \left(\frac{5}{2} R\right) \Delta T \]

4. **Solve for Temperature Change**: Rearranging the equation to find \( \Delta T \):
\[ \Delta T = \frac{2Q}{5nR} \]

5. **Calculate Work Done**: The work done \( W \) during the expansion at constant pressure is given by:
\[ W = P \Delta V \]
Using the ideal gas law \( PV = nRT \), we can express \( P \Delta V \) as:
\[ W = nR \Delta T \]

6. **Substituting for \( \Delta T \)**: Now substitute the expression for \( \Delta T \) into the work done equation:
\[ W = nR \left(\frac{2Q}{5nR}\right) \]

7. **Simplifying the Expression**: The \( nR \) terms cancel out:
\[ W = \frac{2Q}{5} \]

### Final Answer:
The work done in the expansion is:
\[ W = \frac{2}{5} Q \]